一行定理中的第二列可以对齐吗

Question

有没有办法将第二列普遍化析取与下一行定理类环境对齐？

\usepackage{amsmath, amssymb, amsthm}


\newtheoremstyle{exercise}
{\topsep}   % above space
{\topsep}   % below space
{\itshape}  % body font
{0pt}       % indent
{\bfseries} % head font
{}         % head punctuation
{5pt plus 1pt minus 1pt} % HEADSPACE
{}          % CUSTOM-HEAD-SPEC

\theoremstyle{exercise}

\newtheorem{provlinje}{~}
\renewcommand*{\theprovlinje}{\alph{provlinje}}

\begin{document}
    
    
\textit{Prooflines}



\begin{provlinje}
    $\forall x\alpha(x)$
\end{provlinje}
\begin{provlinje}
$\forall x(\alpha(x)\vee\beta(x))$
\end{provlinje}
\begin{provlinje}
$\forall x(\alpha(x)\vee\beta(x)\vee\gamma(x))$
\end{provlinje}
\begin{provlinje}
$\forall x(\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x))$
\end{provlinje}
\begin{provlinje}
$\forall x(\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x))$
\end{provlinje}
\begin{provlinje}
$\forall x(\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x)\vee\zeta(x))$
\end{provlinje}
\begin{provlinje}
$\forall x(\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x)\vee\zeta(x)\vee\eta(x))$
\end{provlinje}
\begin{provlinje}
$\forall x(\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x)\vee\zeta(x)\vee\eta(x)\vee\theta(x))$
\end{provlinje}
\begin{provlinje}
$\forall x(\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x)\vee\zeta(x)\vee\eta(x)\vee\theta(x)\vee\iota(x))$
\end{provlinje}

\end{document} ```

Answer 1

正如我在评论中所说，我会使用enumerate环境。加载enumitem包可提供最大的灵活性。

\documentclass{article}    

\usepackage{enumitem}

\begin{document}

\noindent\textit{Prooflines}
\begin{enumerate}[label=\textbf{\alph*},align=left,ref=\textbf{BLA\alph*}]
\item $\forall x \; \phantom( \alpha(x)$
\item $\forall x \; (\alpha(x)\vee\beta(x))$\label{foo}
\item $\forall x \; (\alpha(x)\vee\beta(x)\vee\gamma(x))$
\item $\forall x \; (\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x))$
\item $\forall x \; (\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x))$
\item $\forall x \; (\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x)\vee\zeta(x))$
\item $\forall x \; (\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x)\vee\zeta(x)\vee\eta(x))$
\item $\forall x \; (\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x)\vee\zeta(x)\vee\eta(x)\vee\theta(x))$
\item $\forall x \; (\alpha(x)\vee\beta(x)\vee\gamma(x)\vee\delta(x)\vee\epsilon(x)\vee\zeta(x)\vee\eta(x)\vee\theta(x)\vee\iota(x))$
\end{enumerate}
See step \ref{foo}.

\end{document}